Suppose that $f(x)$ is continuous on $(0,\infty)$ and that $$\int_0^{\infty} \frac{f(x)}{x} \,dx < \infty.$$ Does it follow that $$\int_0^{\infty}\frac{f(x)^2}{x} \,dx < \infty ?$$
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Do you have more information about the function? Maybe it's value range.... – Anastassis Kapetanakis Aug 26 '18 at 12:41
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You may assume that you have $e^{-x}$ for $x\to \infty$ and at most a power singularity at zero. – user385459 Aug 26 '18 at 12:42
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It’s a real valued function !! – user385459 Aug 26 '18 at 12:43
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1I asked that because that's not true unless we have a specified range of that function!! – Anastassis Kapetanakis Aug 26 '18 at 12:48
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What if $f(x)\ge 0$ for all $x$? The answer given (@Riemann) would not apply. – herb steinberg Aug 26 '18 at 15:18
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$$\int_0^{\infty} \frac{\sin x}{x} \,dx =\frac{\pi}{2},$$ but $\int_0^{\infty} \frac{\sin^2 x}{x} \,dx $ is divergent, because $$\frac{\sin^2 x}{x}=\frac{1-cos 2x}{2x}.$$
Riemann
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